# Full text of "Treatise On Analysis Vol-Ii"

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```362       XV    NORMED ALGEBRAS AND SPECTRAL THEORY

(15.8.6)    (i)    Every self-adjoint idempotent e ^ 0 is the sum of a finite number
of irreducible self-adjoint idempotents belonging to Ae.

(ii)    Every left ideal I ^ {0} contains an irreducible self-adjoint idempotent.

Clearly (ii) is a consequence of (i) and (15.8.5). Hence it is enough to
prove (i). If \\e\\2 < 2, then e must be irreducible, for otherwise we should
have \\e\\2 = HeJI2 + \\e2\\2, where et and e2 are self-adjoint idempotents ^0,
hence \\e\\2 2> 2 by (15.8.3). We shall proceed by induction on the smallest
integer n such that \\e\\2 < n. If e =- el + e2 is reducible, where eA and e2 are
orthogonal self-adjoint idempotents ^0, then el = e±e and e2 = e2e, so that
e^ and e2 lie in Ae; moreover, we have

and similarly ||e2||2 < /? — 1, so that the inductive hypothesis can be applied to
el and e2 . This completes the proof.

A left ideal I in A is said to be minimal if I ^ {0} and if there exists no
nonzero left ideal I' 7* I contained in I. Similarly for minimal right ideals.

(15.8.7) A left ideal I in A is minimal if and only if it is of the form I = Ae,
where e is an irreducible self-adjoint idempotent ^0.

If I is minimal, it contains an irreducible self-adjoint idempotent e ^ 0
(15.8.6). We have Ae e I, and e = e2 e Ae, hence Ae = I. Conversely, let e be
an irreducible self-adjoint idempotent ^0, and let I = Ae. Suppose that I
contains a left ideal I' distinct from {0} and I. Let e' be a self-adjoint idem-
potent 7^0 contained in I' (15.8.5). Then we can write e — el + e2, where
e2 = ee' and e1 = e — ee'. We shall show that e± and e2 are orthogonal self-
adjoint idempotents. Since e' e Ae, we have e' = e'e (15.8.3), hence

e2 = ee'ee' = ee'2 = ee' = e2 ,       ee2 = e2,       e2 e = ee'e = ee' = e2 ,
and therefore

*i*2 = 0 - ^2)^2 = 0,       e2 ex = e2(e - e2) = 0,
e2 = (e- e2)2 = e - e2 = el ;

finally e* = (ee'e)* = ee'e = e2 , and hence ex = e — e2 is also self-adjoint. We
shall obtain a contradiction if we can show that ei and e2 are both nonzero.
If e± = 0, it follows that e == ee' el', hence I = T, contrary to hypothesis. On
the other hand, efe2 = e'eef = e'2 = e' ^ 0, and therefore <?2 T6 0. This com-
pletes the proof., it is said to be irreducible.ary elements of A is bounded.
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